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An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to the earth observer, approaches at 0.82c. What is the velocity of the exploration vehicle relative to the spaceship? Give your answer as a decimal fraction of c, i.e. a β factor.

Answer :

Answer:

Explanation:

Expression for relative velocity

= [tex]\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }[/tex]

= (.54 + .82 )c/ [tex]1+ \frac{.54 \times.82}{1}[/tex]

= 1.36 c / 1.4428

= .94 c

β = .94

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